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# There is a circular garden of Diameter = r. Three points are chosen in

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There is a circular garden of Diameter = r. Three points are chosen in  [#permalink]

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13 Oct 2018, 10:29
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Difficulty:

95% (hard)

Question Stats:

35% (02:31) correct 65% (02:18) wrong based on 34 sessions

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There is a circular garden of Diameter = r. Three points are chosen independently and at random on the boundary of the Garden. What is the approximate probability that none of the three points lies more than a straight-line distance of r/2 away from any other of the three points?

(A) 1/9
(B) 1/12
(C) 1/18
(D) 1/24
(E) 1/27

Weekly Quant Quiz #4 Ques 7

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Re: There is a circular garden of Diameter = r. Three points are chosen in  [#permalink]

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13 Oct 2018, 11:28
2
P(P1) =1 .....as can be placed anywhere in the circle. .......................... ref pic 01
p(P2)= 120/360=1/3 as........ p2 can be at max r(radius) dist from p1. thus forms equilateral trangle at both sides when reference OP1. Thus max possible placement is at 1/3 of 360 deg
..............ref pic 02

For P3 only 2 cases:
case 01 : p1 & p2 at same pt thus p3 at r dist , hence fracton again covers 120 deg out of 360 . Hence p3= 120/360=1/3 ...........ref pic 03
case 02 : when p1 & p2 are not at the same point , then the max arc covered by 2 max distant points is 60. p3=60/360 =1/6 ................ref pic 04

P(P1 P2 P3case01)= 1*1/3*1/3 =1/9
P(P1 P2 P3case01)= 1*1/3*1/6 =1/18........ as these are the 2 extreme cases the actual P is in between.
..................Thus Ans is B.
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Re: There is a circular garden of Diameter = r. Three points are chosen in  [#permalink]

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13 Oct 2018, 10:39
According to me option (C) is correct. Please find the solution.

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Re: There is a circular garden of Diameter = r. Three points are chosen in  [#permalink]

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13 Oct 2018, 23:49
gmatbusters wrote:
There is a circular garden of Diameter = r. Three points are chosen independently and at random on the boundary of the Garden. What is the approximate probability that none of the three points lies more than a straight-line distance of r/2 away from any other of the three points?

(A) 1/9
(B) 1/12
(C) 1/18
(D) 1/24
(E) 1/27

Weekly Quant Quiz #4 Ques 7

I followed the below way -
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Re: There is a circular garden of Diameter = r. Three points are chosen in  [#permalink]

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03 Jun 2019, 01:44
u1983 wrote:
P(P1) =1 .....as can be placed anywhere in the circle. .......................... ref pic 01
p(P2)= 120/360=1/3 as........ p2 can be at max r(radius) dist from p1. thus forms equilateral trangle at both sides when reference OP1. Thus max possible placement is at 1/3 of 360 deg
..............ref pic 02

For P3 only 2 cases:
case 01 : p1 & p2 at same pt thus p3 at r dist , hence fracton again covers 120 deg out of 360 . Hence p3= 120/360=1/3 ...........ref pic 03
case 02 : when p1 & p2 are not at the same point , then the max arc covered by 2 max distant points is 60. p3=60/360 =1/6 ................ref pic 04

P(P1 P2 P3case01)= 1*1/3*1/3 =1/9
P(P1 P2 P3case01)= 1*1/3*1/6 =1/18........ as these are the 2 extreme cases the actual P is in between.
..................Thus Ans is B.

Hey!
Your explanation is crisp.But we get two probability 1/9 and 1/18. We are getting 1/6 after adding both. But do tell me What I'm missing. Looking forward for your reply.

Wishes
Harshdeep
Re: There is a circular garden of Diameter = r. Three points are chosen in   [#permalink] 03 Jun 2019, 01:44
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